Question

A constant retarding force of 50 N is applied to a body of mass 20 kg moving initially with a speed of 15 ms–1. How long does the body take to stop?​

Answer 1

Explanation:

Retarding force, F = –50 N

Mass of the body, m = 20 kg

Initial velocity of the body, u = 15 m/s

Final velocity of the body, v = 0

Using Newton’s second law of motion, the acceleration (a) produced in the body can be calculated as:

F = ma

–50 = 20 × a

∴ a = -50/20 = -2.5 ms-2

Using the first equation of motion, the time (t) taken by the body to come to rest can be calculated as:

v = u + at

∴ t = -u / a = -15 / -2.5 = 6 s

Answer 2

Aɴsᴡᴇʀ:

• Time taken to stop a body = 6 seconds

Gɪᴠᴇɴ:

• Force (F) = 50N
• Mass (m) = 20kg
• Initial velocity (u) = 15m/s
• Final velocity (v) = 0

Nᴇᴇᴅ Tᴏ Fɪɴᴅ:

• Time taken to stop

Solution:

Fᴏʀᴍᴜʟᴀ ᴜsᴇᴅ ʜᴇʀᴇ:

• Force = Mass × Acceleration

Pᴜᴛᴛɪɴɢ ᴛʜᴇ ᴠᴀʟᴜᴇs:

➦ 50 = 20 × a

➦ a = 50/20

➦ a = 2.5m/s²

Due to retardation acceleration of body becomes negative.

Tʜᴇʀᴇғᴏʀᴇ:

• Acceleration of body = -2.5m/s²

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Now,

Bʏ ᴛʜᴇ ғɪʀsᴛ ᴇǫᴜᴀᴛɪᴏɴ ᴏғ ᴍᴏᴛɪᴏɴ:

• v = u + at

Pᴜᴛᴛɪɴɢ ᴛʜᴇ ᴠᴀʟᴜᴇs:

➦ 0 = 15 + (-2.5) × t

➦ 2.5t = 15

➦ Time = 15/2.5

➦ Time = 6 seconds

Tʜᴇʀᴇғᴏʀᴇ:

• Time taken to stop a body = 6 seconds

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